Chapter 5: Problem 19
Evaluate the sums. $$ \text { a. }sum_{k=1}^{10} k \quad \text { b. } \sum_{k=1}^{10} k^{2} \quad \text { c. } \sum_{k=1}^{10} k^{3} $$
Short Answer
Expert verified
a) 55, b) 385, c) 3025.
Step by step solution
01
Understanding the Problem
We need to evaluate three different sums involving powers of integers from 1 to 10. These are: a) sum of integers, b) sum of squares of integers, and c) sum of cubes of integers.
02
Solve Part a
The sum of the first 10 positive integers can be calculated using the formula for the sum of an arithmetic series: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]. Applying this to \( n = 10 \), we get: \[ \sum_{k=1}^{10} k = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55 \].
03
Solve Part b
The sum of the squares of the first 10 positive integers can be calculated using the formula: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]. Applying this formula for \( n = 10 \), yields: \[ \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 385 \].
04
Solve Part c
The sum of the cubes of the first 10 positive integers is given by the formula: \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]. For \( n = 10 \), we calculate: \[ \sum_{k=1}^{10} k^3 = \left(\frac{10 \times 11}{2}\right)^2 = 55^2 = 3025 \].
05
Summarize Results
We have evaluated each part of the sums separately. For part a, the sum is 55; for part b, it is 385; and for part c, the sum is 3025.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arithmetic Series
An arithmetic series is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the previous term. The series can be written as \( a, a+d, a+2d, a+3d, \ldots \), where \( a \) is the first term and \( d \) is the common difference.
The sum of the first \( n \) terms of an arithmetic series can be calculated using the formula:
The sum of the first \( n \) terms of an arithmetic series can be calculated using the formula:
- \( S_n = \frac{n}{2} (2a + (n-1)d) \)
- \( S_n = \frac{n(n+1)}{2} \)
- \( S_{10} = \frac{10 imes 11}{2} = 55 \)
Sum of Squares
The sum of squares is concerned with adding together the squares of a series of numbers, such as \( 1^2, 2^2, 3^2, \ldots, n^2 \). This measure is crucial in various mathematical contexts, such as statistics and physics.
The formula to calculate the sum of squares of the first \( n \) natural numbers is:
The formula to calculate the sum of squares of the first \( n \) natural numbers is:
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10 + 1)}{6} = 385 \)
Sum of Cubes
The sum of cubes is the total of each number cubed from 1 to \( n \). This concept is often used in algebra and number theory to discover interesting properties of numbers.
The formula to determine the sum of cubes is an elegant relation:
The formula to determine the sum of cubes is an elegant relation:
- \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
- For example, \( \sum_{k=1}^{10} k^3 = 55^2 = 3025 \)